We will have that the deceleration can be determined as follows:
First, we recall that for linear accelerated motions the following is true:
[tex]v_f^2=v_i^2+2ad[/tex]So:
[tex]\begin{gathered} \Rightarrow(0m/s)^2=(13.41m/s)^2+2a(60m)\Rightarrow2a(60m)=-(13.41m/s)^2 \\ \Rightarrow a=\frac{-(13.41m/s)^2}{2(60m)}\Rightarrow a=-1.4985675...m/s^2 \\ \\ \Rightarrow a\approx-1.5m/s^2 \end{gathered}[/tex]So, the deceleration must be approximately of 1.5 m/s^2.
